在图像中查找强度相同的项目

Question

How can I find all the points (or the region) in an image which has 3-dimensions in which the first two dimensions show the resolution and the 3rd one shows the density? I can use Matlab or Python. I wonder if there is a native function for finding those points that is least computationally expensive. UPDATE: Imagine I have the following:

A= [1,2,3; 4,6,6; 7,6,6]
A =

     1     2     3
     4     6     6
     7     6     6

>> B=[7,8,9; 10,11,11; 1, 11,11]

B =

     7     8     9
    10    11    11
     1    11    11

>> C=[0,1,2; 3, 7, 7; 5,7,7]

C =

     0     1     2
     3     7     7
     5     7     7

How can I find the lower square in which all the values of A equal the same all the values of B and all the values of C? If this is too much how can I find the lower square in A wherein all the values in A are equal? *The shown values are the intensity of the image.

UPDATE: tries the provided answer and got this error:

>> c=conv2(M,T, 'full');
Warning: CONV2 on values of class UINT8 is obsolete.
         Use CONV2(DOUBLE(A),DOUBLE(B)) or CONV2(SINGLE(A),SINGLE(B)) instead. 
> In uint8/conv2 (line 10) 
Undefined function 'conv2' for input arguments of type 'double' and attributes 'full 3d real'.

Error in uint8/conv2 (line 17)
y = conv2(varargin{:});

*Also tried convn and it took forever so I just stopped it! Basically how to do this for a 2D array as described above?

Answer 1

You can use a pair of horizontal and vertical 1D filters such that the horizontal filter has a kernel of [1 -1] while the vertical filter has a kernel of [1; -1] [1; -1] . The effect of this is that it takes both horizontal and vertical pairwise distances for each element in each dimension separately. You can then perform image filtering or convolution using these two kernels ensuring that you replicate the borders. To be able to find uniform regions, by checking which regions in both results map to 0 between them both, this gives you areas where areas that are uniform over all channels independently.

To do this, you would first take the opposite of both filtering results so that uniform regions that would become 0 are now 1 and vice-versa. that you perform the logical AND operation on both of these together and then ensure that for each pixel temporally, all of the values are true . This would mean that for a spatial location in this image, all values experience the same uniformity as you expect.

In MATLAB, assuming you have the Image Processing Toolbox, use imfilter to filter the images, then use all in MATLAB to look temporally after the two filtering results, and then use regionprops to find the coordinates of the regions you seek. So do something like this:

%# Reproducing your data
A = [1,2,3; 4,6,6; 7,6,6];
B = [7,8,9; 10,11,11; 1, 11,11];
C = [0,1,2; 3, 7, 7; 5,7,7];

%# Create a 3D matrix to allow for efficient filtering
D = cat(3, A, B, C);

%# Filter using the kernels
ker = [1 -1];
ker2 = ker.'; %#
out = imfilter(D, ker, 'replicate');
out2 = imfilter(D, ker2, 'replicate');

%# Find uniform regions
regions = all(~out & ~out2, 3);

%# Determine the locations of the uniform areas
R = regionprops(regions, 'BoundingBox');

%# Round to ensure pixel accuracy and reshape into a matrix
coords = round(reshape([R.BoundingBox], 4, [])).';

---

coords would be a N x 4 matrix with each row telling the upper-left coordinates of the bounding box origin as well as the width and height of the bounding box. The first and second elements in a row are the column and row coordinate while the third and fourth elements are the width and height of the bounding box.

The regions we have detected can be found in the regions variable. Both of these show:

>> regions

regions =

  3×3 logical array

   0   0   0
   0   1   1
   0   1   1

>> coords

coords =

     2     2     2     2

This tells us that we have localised the region of "uniformity" to be the bottom right corner while the coordinates of the top-left corner of the bounding box are row 2, column 2 with a width and height of 2 and 2 respectively.

Answer 2

A possible solution:

A = [1,2,3; 4,6,6; 7,6,6];
B = [7,8,9; 10,11,11; 1, 11,11];
C = [0,1,2; 3, 7, 7; 5,7,7];
%create a 3D array
D = cat(3,A,B,C)
%reshape the 3D array to 2D 
%its columns represent the third dimension
%and its rows represent resolution
E = reshape(D,[],size(D,3));
%third output of the unique function applied row-wise  to the data
%represents the label of each pixel a [m*n, 1] vector created
[~,~,F] = unique(E,'rows');
%reshape the vector to a [m, n] matrix of labels
result = reshape(F, size(D,1), size(D,2));

You can reshape the 3D matrix to a 2D matrix ( E ) that its columns represent the third dimension and its rows represent resolution.

Then using unique function you can label the image.

We have a 3D matrix:

A =
     1     2     3
     4     6     6
     7     6     6
B =
     7     8     9
    10    11    11
     1    11    11
C =
     0     1     2
     3     7     7
     5     7     7

When we reshape the 3D matrix to a 2D matrix E we get:

E =

    1    7    0
    4   10    3
    7    1    5
    2    8    1
    6   11    7
    6   11    7
    3    9    2
    6   11    7
    6   11    7

So we need to classify the rows base on their values.

Unique function is capable of extracting unique rows and assign the same label to rows that are equal to each other.

Here varible F capture third output of the unique function that is label of each row.

F =

   1
   4
   6
   2
   5
   5
   3
   5
   5

that should be reshaped to 2D

result =

1   2   3
4   5   5
6   5   5

so each region has different label.

If you want to segment distinct regions(based on both their values and their spatial positions) you need to do labeling the image in a loop

numcolors = max(F);
N = 0;
segment = zeros(size(result));
for c = 1 : numcolors
    [label,n] = bwlabel(result==c);
    segment = segment +label + logical(label)*N;
    N = N + n;
end

So here you need to mark disconnected regions that have the same values with different labels. since MATLAB doesn't have functions for gray segmentation You can use bwlabel function multiple times to do segmentation and add result of the previous iteration to result of current iteration. segment variable contains the segmentd image.

*Note: this result obtained from GNU Octave that its labeling is different from MATLAB. if You use unique(E,'rows','last'); result of MATLAB and Octave will be the same.

Answer 3

check out https://docs.scipy.org/doc/scipy-0.15.1/reference/generated/scipy.signal.correlate2d.html

2D correlation basically "slides" the two images across each other, and adds up the dot product of the overlap.

more reading: http://www.cs.umd.edu/~djacobs/CMSC426/Convolution.pdf

https://en.wikipedia.org/wiki/Two-dimensional_correlation_analysis